If $y = {x^x}$, prove that $\dfrac{{{d^2}y}}{{d{x^2}}} - \dfrac{1}{y}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} - \dfrac{y}{x} = 0$ .
Answer
Verified361.2k+ views
Hint: In this question first we take log both side and simply differentiate expression $y = {x^x}$ with respect to $x$ and find the value of $\dfrac{{dy}}{{dx}}$ & $\dfrac{{{d^2}y}}{{d{x^2}}}$ after that value of $\dfrac{{dy}}{{dx}}$ &$\dfrac{{{d^2}y}}{{d{x^2}}}$ and put left hand side of differential expression $\dfrac{{{d^2}y}}{{d{x^2}}} - \dfrac{1}{y}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} - \dfrac{y}{x} = 0$.
Let $\dfrac{{{d^2}y}}{{d{x^2}}} - \dfrac{1}{y}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} - \dfrac{y}{x} = 0$ ……. (1)
Consider $y = {x^x}$
Take log both side
$\log y = \log {x^x}$
Apply log property $\left( {\log {a^b} = b\log a} \right)$
$\log y = x\log x$
Differentiate with respect to $x$
$\dfrac{1}{y}\dfrac{{dy}}{{dx}} = x \times \dfrac{1}{x} + \log x$
$\dfrac{{dy}}{{dx}} = y\left( {1 + \log x} \right)$ ……. (2)
Differentiate equation (2) with respect to $x$
$\dfrac{{{d^2}y}}{{d{x^2}}} = y\left( {0 + \dfrac{1}{x}} \right) + \left( {1 + \log x} \right)\dfrac{{dy}}{{dx}}$
Put value of $\dfrac{{dy}}{{dx}}$ in above expression
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{y}{x} + y{\left( {1 + \log x} \right)^2}$…….(3)
Value of $\dfrac{{dy}}{{dx}}$ and $\dfrac{{{d^2}y}}{{d{x^2}}}$ from equation 2 and 3 put into equation 1
$\dfrac{y}{x} + y{\left( {1 + \log x} \right)^2} - \dfrac{1}{y}{\left( {y\left( {1 + \log x} \right)} \right)^2} - \dfrac{y}{x} = 0$
$\dfrac{y}{x} + y{\left( {1 + \log x} \right)^2} - y{\left( {1 + \log x} \right)^2} - \dfrac{y}{x} = 0$
Here we can see that after solving left hand side will be zero
So L.H.S=R.H.S
Hence proved
Note: In this question we use log property $\log {a^b} = b\log a$ and also we use some basic differentiation formula like
$
\dfrac{d}{{dx}}\log x = \dfrac{1}{x} \\
\\
$
$\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}v + v\dfrac{d}{{dx}}u$ here $u$ and $v$ are function of real variable $x$ and this formula also known as product rule for derivatives.
$\dfrac{d}{{dx}}x = 1$
Let $\dfrac{{{d^2}y}}{{d{x^2}}} - \dfrac{1}{y}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} - \dfrac{y}{x} = 0$ ……. (1)
Consider $y = {x^x}$
Take log both side
$\log y = \log {x^x}$
Apply log property $\left( {\log {a^b} = b\log a} \right)$
$\log y = x\log x$
Differentiate with respect to $x$
$\dfrac{1}{y}\dfrac{{dy}}{{dx}} = x \times \dfrac{1}{x} + \log x$
$\dfrac{{dy}}{{dx}} = y\left( {1 + \log x} \right)$ ……. (2)
Differentiate equation (2) with respect to $x$
$\dfrac{{{d^2}y}}{{d{x^2}}} = y\left( {0 + \dfrac{1}{x}} \right) + \left( {1 + \log x} \right)\dfrac{{dy}}{{dx}}$
Put value of $\dfrac{{dy}}{{dx}}$ in above expression
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{y}{x} + y{\left( {1 + \log x} \right)^2}$…….(3)
Value of $\dfrac{{dy}}{{dx}}$ and $\dfrac{{{d^2}y}}{{d{x^2}}}$ from equation 2 and 3 put into equation 1
$\dfrac{y}{x} + y{\left( {1 + \log x} \right)^2} - \dfrac{1}{y}{\left( {y\left( {1 + \log x} \right)} \right)^2} - \dfrac{y}{x} = 0$
$\dfrac{y}{x} + y{\left( {1 + \log x} \right)^2} - y{\left( {1 + \log x} \right)^2} - \dfrac{y}{x} = 0$
Here we can see that after solving left hand side will be zero
So L.H.S=R.H.S
Hence proved
Note: In this question we use log property $\log {a^b} = b\log a$ and also we use some basic differentiation formula like
$
\dfrac{d}{{dx}}\log x = \dfrac{1}{x} \\
\\
$
$\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}v + v\dfrac{d}{{dx}}u$ here $u$ and $v$ are function of real variable $x$ and this formula also known as product rule for derivatives.
$\dfrac{d}{{dx}}x = 1$
Last updated date: 19th Sep 2023
•
Total views: 361.2k
•
Views today: 10.61k
Recently Updated Pages
